Euler - Perfect cube question

[Averagepunter]

What is the smallest natural number n such that n is not a perfect cube, but the number of factors of n³ is a perfect cube?

 

[mmm4444bot]

Hello Averagepunter. Is this exercise part of a school assignment? What have you done so far?

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[pka]

What is the smallest natural number n such that n is not a perfect cube, but the number of factors of n³ is a perfect cube?

How many divisors (factors) does \(\displaystyle 30^3\) have?

 

[Steven G]

Have you tried to see if it works for n=1, n=2, n=3. ...? Sometimes it is good to do a problem that way, get the answer and then think of a more elegant way of doing it. If you don;t see an elegant then at least you have the answer.

 

[Averagepunter]

How many divisors (factors) does \(\displaystyle 30^3\) have?

Thank you very much ! n = 30 or 30³ has 64 factors and 64 is a perfect cube ! Thanks for your help. Much appreciate it.

 

[Averagepunter]

Have you tried to see if it works for n=1, n=2, n=3. ...? Sometimes it is good to do a problem that way, get the answer and then think of a more elegant way of doing it. If you don;t see an elegant then at least you have the answer.

Thank you for your help. After I tried 30 x 30 x 30 as suggested you and pka; I got 27000, which is (27 x 1000). 27 has 4 factors and 1000 has 16 factors. So, (4 x 16) = 64 factors? Is that the elegant way to find out the solution?

 

[pka]

Thank you for your help. After I tried 30 x 30 x 30 as suggested you and pka; I got 27000, which is (27 x 1000). 27 has 4 factors and 1000 has 16 factors. So, (4 x 16) = 64 factors? Is that the elegant way to find out the solution?

Do you know the way to count the number of divisors a positive integer \(\displaystyle N\) has?
First factor \(\displaystyle N=p_1^a\cdot p_2^b\cdot p_3^c\cdots p_n^m\) where each of the \(\displaystyle p_i\) is a prime factor with a positive integer exponent.
Then \(\displaystyle N\) has \(\displaystyle (a+1)\cdot(b+1)\cdot(c+1)\cdots(m+1)\) divisors.

Example: If \(\displaystyle N=2^3\cdot 5^2\cdot 7^5\cdot 13^1\cdot 19^4\) then \(\displaystyle N\) has \(\displaystyle (3+1)\cdot (2+1)\cdot (5+1)\cdot (1+1)\cdot (4+1)
Look at this LINK. Notice that WolfFram gives the factorization & the divisor count.
You need to ask yourself: "Why add \(\displaystyle 1\) to each exponent?" (HINT: \(\displaystyle 1\) is a divisor).\)

 

[Otis]

… Is [my explanation] the elegant way to find out the solution?

Maybe pka will provide the answer to that question, too.

 

[Steven G]

Thank you for your help. After I tried 30 x 30 x 30 as suggested you and pka; I got 27000, which is (27 x 1000). 27 has 4 factors and 1000 has 16 factors. So, (4 x 16) = 64 factors? Is that the elegant way to find out the solution?

How do you know that the number pka is the smallest number satisfying your condition? Just because pka stated a number you automatically think it is the number that you are looking for. Is that real mathematics?

 

[Averagepunter]

How do you know that the number pka is the smallest number satisfying your condition? Just because pka stated a number you automatically think it is the number that you are looking for. Is that real mathematics?

Well, I started of hard way by doing cubes of numbers starting from 1, 2, 3 and so on to find out a number whose cube has factors equating to either 8, 27, 64 etc. (Perfect cube). I persisted up to about 23 or 24 and then got tired. So, trying up to 30 made sense.
Please enlighten me, if you have a better way to find it out? your "elegant" way. I would only be too grateful if you shared it.
Thank you in advance.